Measured radius with a small sector

[[Di...]]

Hello! I have a problem: measure the lens (radius 4000 mm) , height sector of lens 25 mm. I use a function in Calypso "radius by point". I have result about 3999.9997 mm, but form 0.1567 mm . I think , it false. Do you know other way measurement a small of sector.(foto other lens for example)
Thanks)

IMG_20190411_185656.jpg

[[Kl...]]

Hey
a center and diameter or radius is only mathematical correct calculated if the circle are nearly closed.
Zero Circle
x^2 + Y^2 = R^2
For segments the Formular
x(R)^2 + y(R)^2 = R^2
means the form deviation has a parallel influence on center and radius.
In direction of the open area the center is normaly shifted.

Most Software offers therefore two possibilities, additional Curve
- Fix the Center by theoretical values from drawing and check the form deviation, must be very small, by using a hard outlier elimimaton and filter
- Fix the radius and calculate the center. Form deviation must be also very small
- Last check it by curve with nominal created by drawing values. If deviations are inside the tolerance everthing is OK. Filter and outlier see above.

Sorry there are no simple methode to get a statement

Klaus-Dieter

[[Er...]]

Not to start a shit storm, but what you say is not true.
You could have 3 points on a arc segment of 0.1°. It would still be mathematically correct calculated. It kind of depend on the number of decimals where the center will end up. And in this case the decimals corresponds to cmm inaccuracy and part nonconformance.

p1=point(1.1368683772162e-13,1898.5,0)
p2=point(-16.567327643663,1898.4277109373,0)
p3=point(-33.133393621183,1898.2108492544,0)

x1=p1.x
y1=p1.y

x2=p2.x
y2=p2.y

x3=p3.x
y3=p3.y

temp = (x2 * x2) + (y2 * y2)
bc = (x1 * x1 + y1 * y1 - temp) / 2
cd = (temp - x3 * x3 - y3 * y3) / 2
det = (x1 - x2) * (y2 - y3) - (x2 - x3) * (y1 - y2)

if abs(det) < 0
    message("Ingen beräkning kan ske då punkterna bildar en linje"+cr()+"Allvarligt fel")
else
	cx = (bc*(y2 - y3) - cd*(y1 - y2)) / det
	cy = ((x1 - x2) * cd - (x2 - x3) * bc) / det
	display("Center: "+cx+" "+cy)
	display("Radius: "+sqrt((cx - x1)**2 + (cy - y1)**2))
endif

I think what you are trying to say is that the accuracy of a such measurement would no be very good due to the form error making the point coordinates spread from a common center. I guess it's here fitting algorithms comes in, but that's another dark path to take another time.

But I don't fully grip how your pythoragos has anything to do with it, if the circle are closed (lets say 3points, 1 point per 180°)
or not, more than showing the 2d deviation from nominal. But that's not the way to describe this "CMM short arc problem".

But I'm open for learning 🙂

[[Di...]]

Thanks my colleagues for ideas. 🙂

Please sign in to view this quote.

You are right. I search a true way, I know it don't simple method.
I fix the Center by theoretical values from drawing and check the form deviation, was very bed, l used fillter.
A good idea with curve, but l don't have cad-model. I will be think)))

[[Er...]]

If you are only trying to confirm the radius, I think space points is a quite reliable method. (Threat it as a freeform surface.)
But if you are trying to figure out where the focal point of the lens ends up or more advance stuff. I don't really have any good idea.

As long as you know the nominal radius/radiuses (plural?) you can just create a sphere and fill it up with points. I personally don't believe you will be successful with the curve approach.

[[Kl...]]

You've right Eric,
- with 3 points always a exact circle can be calculated
- space point is a fast method to verify the radius at local points
- the form deviation has an influence, if multiple points are used (scanning)

When last method is used the correlation between radius and center takes place

Inside the ..\inspection\02_PDF the Table_Trend.pdf shows this relation
=> Z_Value and diameter deviation are shifting nearly parallel, while X_Value is nearly constant
=> clamping the center, the diameter will be nearly constant.

Results created by dispersion from 0.0 to ±0.0020 by step 0.0001, 5000 measured points, FilterLambda = 2.5,
Segment angle 2.5°, AngleStep Point Generator = 0.002°, iterativ outlier elimination factor = 5

This test dosn't shows the effect of the segment angle (180° ==> 2.5°), gradient variation of the lens segment or tilted shape of the lens (influence on X).

To find the center of the radius will be always critical, because this has an influence one all calculations circle, curve or space point

Dina,
inside the inspection you found the method to generate a curve without CAD by using the Point Generator. I prefer this, sometimes the CAD are not exact enough, and I will get the maximum number of decimal places (no rounding during calculation only for protocol output).

The standard curve calcutation runs to a limit, if the height becomes very small. Bearing manufucturer uses a Special mathematic (logarithmic curve calculation).

Klaus-Dieter

Lens.zip

[[Er...]]

And I agree with your demonstration. I just dissagreed with the statement:

Please sign in to view this quote.

This isnt a mathematical problem. How ever, if you have 4 point (to keep it easy) and the they dont have a common center. Then is not by definition a circle. But this is not a mathematical "problem" either, since there are several methods to average/mean (as an example) out a radius/center from the points you have.

The biggest problem in this case is always the quality of the data. 🙂